KayG12
Answered

What is the equation of the circle with center (-6,7) that passes through the point (4,-2). A. (x+6)^2+(y-7)^2=181 B. (x-6)^2+(y+7)^2=181 C. (x+6)^2+(y-7)^2=9 D. (x-6)^2+(y+7)^2=9

Answer :

apologiabiology
the equation of a cricle passing through (h,h) and with radius r is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
so

cente ris (-6,7) and passing through (4,-2)
[tex](x-(-6))^2+(y-7)^2=r^2[/tex]
[tex](x+6)^2+(y-7)^2=r^2[/tex]
subsitute (4,-2) to find r^2

[tex](4+6)^2+(-2-7)^2=r^2[/tex]
[tex](10)^2+(-9)^2=r^2[/tex]
[tex]100+81=r^2[/tex]
[tex]181=r^2[/tex]

so it is

[tex] (x+6)^2+(y-7)^2=181[/tex]
A is the answer

Other Questions